New answers tagged

0

I'd use Millman's theorem to solve for voltage \$\color{red}{V_X}\$: - The theorem says this (DC version with resistors): - But it also applies to AC and complex impedances hence: - $$V_X = \dfrac{\dfrac{V_I}{R1}+\dfrac{0}{R2 + \frac{1}{sC1}}+\dfrac{V_O}{\frac{1}{sC2}}}{\dfrac{1}{R1} +\dfrac{1}{R2 + \frac{1}{sC1}} +\dfrac{1}{\frac{1}{sC2}}}$$ And, of ...


5

This circuit is a 7th-order system and if you apply the impedance divider as recommended, then you'll end up in a total algebraic paralysis unless you want to use a computer for doing it. If you want to derive by hand and express it in a readable polynomial format, the best is to apply the fast analytical circuits techniques or FACTs, they are unbeatable in ...


1

... my professor keeps telling me that I need to use the formula \$H = \lvert \frac{Z_2}{Z_1 +Z_2}\rvert\$ This is what's called a ladder filter. There's a series element, then a shunt element, then series, then shunt and so on, from one end to the other. That equation of your Prof is fine, but you'll have to iterate it for each series/shunt pair in the ...


1

A higher sampling frequency means more bandwidth below the nyquist point (the right hand end of the X-axis). So can we say it doesn't change the transfer function correct? As your sampling rate approaches the cut-off frequency then the characteristic is highly affected due to a thing called frequency warping. I see, but if the sampling rate is beyond the ...


1

It is not changing the filter cut-off frequency. However, the x axis is from 0 to fs/2. As you vary the sampling frequency the x axis limits effectively change and the cut off frequency location within the graph changes. It doesn't change the transfer function ? It does. The filter coefficients as well as the response. Below is my result in Octave software ...


0

I don't know the answer directly, but if the signal power levels aren't too high, I'd be tempted to just try to design this with discrete components: I don't promise it will be easy to tune the component values in the real world to get the performance you want.


3

But I can't see this as an accepted solution anywhere. Is there any good reason for that? This is a very common solution and most people have probably encountered a ferrite clamp on a cable at some point in their life. Many cables use a ferrite clamp to reduce high frequency noise on the shield. High frequency currents take the path of least inductance, so ...


0

The supply is supposed to be steady, or having small variations, so the only major events should be turn on and turn off. Then, the snubber acts as a differentiator for the voltage across it, so the main factor contributing to the current will be the variation of the voltage, or (in the case of on/off) the rising and falling times. Which means that the ...


2

D1 in that circuit is a Transient Voltage Suppressor, this is designed to protect other devices attached between VCC and GND from short term over voltage events. It doesn't really provide any filtering from noise though. If it burned that suggests it was subjected to a longer lasting period of over voltage than it was able to handle.


-1

I was asked to delete my obvious ways to design a circuit to do it, since this is homework and there is no actual teacher to help the student.


1

greatly appreciate any pointer! It appears (and it isn't that obvious so I might be wrong) that you have an input filter circuit that is very highly tuned as per this in your question: - And if that is so, then what you are seeing (when data changes the FSK frequency from 23.5 kHz to 32.2 kHz) is the tuned circuit "ringing" for far too long. I ...


0

This transfer function can be solved without writing a single line of algebra while going straight to the point with a low-entropy expression. I will be using the fast analytical circuits techniques or FACTs as described in the book I published in 2016. The principle is truly simple and applies well to passive circuits of any order: determine the time ...


0

Eventually, I figured it out. All the equations at the original post are correct. However, I had to express the inductors' currents \$ I_1, I_2 \$ as a function of duty cycle \$ D \$, input voltage \$ V_g \$ and load resistor \$ R \$. To sum everything up: $$ V = DV_g $$ $$ I_2 = \frac{V}{R} \Rightarrow I_2 = \frac{DV_g}{R} $$ $$ I_1 = DI_2 \Rightarrow I_1 = ...


1

You have positive feedback in your circuit. The output is latched high. The reason for the fall off at higher frequencies is the internal behaviour of the op-amp as you've mentioned.


1

Solved: When applying the complementary transformation I forgot to swap my infinity resistance r1 and my zero resistance r2. The positive terminal of the OPAMP should be connected to ground.


0

You have found the sensitivities. But, now you need to use the actual expected variation in the components and the GBW to compute the variation in the response. For each component that varies, multiply the variation of the component by the sensitivity. Add up the contribution of all the components in a square root of the sum of the squares fashion. Then ...


1

It's a second order low pass filter so use a signal generator and oscilloscope and look for the magnitude peak and frequency where it occurs: - The peak amplitude can be converted to zeta (\$\zeta\$) and zeta (aka the damping ratio) relates to Q-factor as follows: - $$Q = \dfrac{1}{2\zeta}$$ And you can plug \$\zeta\$ into the peaking frequency formula to ...


0

Because you're a user of Mathematica I will use it as well. Now we can find the magnitude response by using the following code: In[1]:=R1 = 10*1000; R2 = 10*1000; C1 = 150*10^(-9); C2 = (33/20)*10^(-9); s = \[Omega]*I; \[Omega] = 2*Pi*f; x = (1/(R1*R2*C1*C2))/(s^2 + ((1/(R1*C1)) + (1/(R2*C1)))* s + (1/(R1*R2*C1*C2))); FullSimplify[ Sqrt[(ComplexExpand[...


0

Solution B doesn't really do much. You have both sides connected to the same ground, so it isn't really isolating. If it is removing glitches, then it is because of some side effect of the opto-isolator. If you change opto-isolators or the noise source changes, then noise may get through again and you won't have many options to fix the problem. I'd stick ...


0

Because both sides of the opto share a common GND, it is not addressing one of the most common noise sources. However, it does add something to the circuit. Basically, it is functioning as a comparator without hysteresis. And, significantly, without debounce. You can debounce the switch signal in software, but you have not stated that this is being done. ...


-1

A) You are correct. B) Yet again correct. My preference would be the optocoupler. Although it adds an extra component to the BOM, the overal component count is lower. The issue with having to choose a potiential different optocoupler for a different frequency should not be a big issue. Many optocouplers (as far a quick browse showed me) use the same pinout. ...


1

There is one way to get this transfer function right without pain and multiple iterations: use the fast analytical circuits techniques or FACTs as described in my book. By determining the time constants of this circuits with a zeroed excitation when the energy-storing elements are set in their dc or high-frequency states, the transfer function comes easily ...


0

Try this: - Image from here. Basically, a half wave transmission line acts like a series tuned circuit and, when resonant, it will act as a direct connection between input and output (just like a series tuned circuit at resonance).


4

The effect as mentioned by Spehro Pefhany is even more severe because of the following: For rising frequencies the capacitive impedance in the feedback path gets lower and lower - and an increasing portion of the input signal arrives DIRECTLY via this feedback capacitor at the opamps (increasing) output impedance. This works against the "normal" ...


9

This is a "well-known" issue with the Sallen-Key configuration. The problem is that the output impedance of the amplifier rises with frequency and starts to look inductive so it's no longer 'fighting' the input signal more as the frequency increases. The output then increases by +20dB/decade until it levels out at the GBW of the amplifier. You can ...


0

I think that you can use an active filter topology - however, not the positive Sallen-Key structure as shown in your description of the problem. The reason is as follows: With rising frequencies, the open-loop gain of the opamp decreases (this alone is NOT already the problem), but the loop gain as well as the feedback effect are also decreasing. This means, ...


2

The awful truth is S&K LPF has problems with suppressing the HF spectrum of the square edge is f-3dB=0.35/tR is far greater than the 1MHz or so GBW of the Op Amp. So a spike passes thru and then gets integrated and then held by the low impedance DC output. Negative feedback is unable to deal effectively with a fast square wave input. Solution? Don't ...


1

If it's about calculating the bandwidth, it would be better to calculate the transfer function instead of relying on pre-made formulas: $$\begin{align} Z_{ser}(s)&=R_{ser}+sL\tag{1} \\ Z_{par}(s)&=\frac{1}{\frac{1}{Z_{ser}}+sC}\tag{2} \\ H(s)&=\frac{R}{R+Z_{par}(s)} \\ &\Rightarrow \\ H(s)&=\frac{RLCs^2+RR_{ser}Cs+R}{RLCs^2+(RR_{ser}C+L)s+...


1

All of these components are nominally 50 Ohms. For a receiver, improving the impedance match doesn't have a big payoff, so if you're not experienced, you are probably better off putting a DC blocking capacitor between the each stage and moving on. I'd suggest this, because the LNA may be biased internally in such a way as to have a DC voltage at either the ...


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